Question

A sample mean, sample standard deviation, and sample size are given. Use the one mean t-test to perform the required hypothesis test about the mean M of the population from which the sample was drawn. Use the P- value approach. Also asses the strength of the evidence against the null hypothesis. Mean of the sample= 226,450 ; S= 11,500 ; n= 23 ; mean of the population = 220, 000 ; Ha : M>220,000 ; alpha = 0.01 .

2). The engendering school at a major university claim that 25% of its graduate are women. In graduating class of 210 students, 43 were females. Does this suggest that the school is believable? Use alpha = 0.10.

Answer 1

1)

Ho :   µ =   220000  
Ha :   µ >   220000   (Right tail test)
          
Level of Significance ,    α =    0.010  
sample std dev ,    s =    11500.0000  
Sample Size ,   n =    23  
Sample Mean,    x̅ =   226450.0000  
          
degree of freedom=   DF=n-1=   22  
          
Standard Error , SE = s/√n =   11500/√23=   2397.9158  
t-test statistic= (x̅ - µ )/SE =    (226450-220000)/2397.9158=   2.690  
          
          
p-Value   =   0.0067   [Excel formula =t.dist(t-stat,df) ]
Decision:   p-value≤α, Reject null hypothesis       
Conclusion: There is enough evidence that mean is greater than population mean.

2)

Ho :   p =    0.25  
H1 :   p ╪   0.25   (Two tail test)
          
Level of Significance,   α =    0.10  
Number of Items of Interest,   x =   43  
Sample Size,   n =    210  
          
Sample Proportion ,    p̂ = x/n =    0.2048  
          
Standard Error ,    SE = √( p(1-p)/n ) =    0.0299  
Z Test Statistic = ( p̂-p)/SE =    (0.2048-0.25)/0.0299=   -1.51  
          
          
p-Value   =   0.1300   [excel formula =2*NORMSDIST(z)]
Decision:   p value>α ,do not reject null hypothesis       
There is not enough evidence to conclude that the school is believable.

  

Please let me know in case of any doubt.

Thanks in advance!

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